## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 683: 107

#### Answer

The required solution is $\frac{\sqrt{3}}{2}$

#### Work Step by Step

Let, \begin{align} & {{\sin }^{-1}}\frac{\sqrt{3}}{2}=x \\ & \sin x=\frac{\sqrt{3}}{2} \\ \end{align} The value of $\sin \frac{\pi }{3}$ is $\frac{\sqrt{3}}{2}$. Thus, \begin{align} & \sin x=\sin \frac{\pi }{3}\text{ } \\ & x=\frac{\pi }{3} \end{align} The value of ${{\sin }^{-1}}\frac{\sqrt{3}}{2}=x$, and $x=\frac{\pi }{3}$ . Therefore, the value of ${{\sin }^{-1}}\frac{\sqrt{3}}{2}$ is $\frac{\pi }{3}$. Thus, the expression is further simplified by putting the value of ${{\sin }^{-1}}\frac{\sqrt{3}}{2}$. \begin{align} & \sin \left( 2{{\sin }^{-1}}\frac{\sqrt{3}}{2} \right)=\sin \left( 2\cdot \frac{\pi }{3} \right) \\ & =\sin \left( \frac{2\pi }{3} \right)\text{ } \\ & =\frac{\sqrt{3}}{2} \end{align}

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